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\title{Ramazan Kasim Arikan - 2008719012}

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setting
 $y=ax+b$ in (1) we have \\
 \\
$(a^2-1)x^3+(-2a^2+2ab-2)x^2+(b^2-4ab)x-2b^2=0$\\

$\Rightarrow a=\pm1\Rightarrow -2\pm2b-2=0 \Rightarrow b=\pm2$ \\
\\
$y=x+2$ and $y=-x-2$ are the inclined asymptotes. \\

\begin{exmp}
Sketch the curve of the algebraic function \\
\\
$y=f(x)=x\sqrt\frac{x+2}{x-2}$ \\
\\
\underline {Solution}.
$D\tiny{f}=R-(-2,-2]$ \\
\\
There is one vertical asymptote at x=2.\\

In Example 1 asymptotes were obtained by the use of a Theorem. We obtain here the oblique (and horizontal) asymptote\\
$y=ax+b$, by direct use of limits:
\\

$a=\displaystyle\lim_{x\to\infty}\frac{f(x)}{x}=
   \displaystyle\lim_{x\to\infty}\sqrt\frac{x+2}{x-2}=1$

$b=\displaystyle\lim_{x\to\infty}[f(x)-x]=
   \displaystyle\lim_{x\to\infty}[\sqrt\frac{x+2}{x-t}-1]$

   $=\displaystyle\lim_{x\to\infty}\frac{\left(\frac{x+2}{x-2}\right)^{1/2}-1}{1/x}
=\displaystyle\lim_{x\to\infty}\frac{\frac{1}{2}\left(\frac{x+2}{x-2}\right)^{-1/2}\frac{-4}{(x-2)^2}}{-1/x^2}=2$
\\
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Oblique asymptote: y=x+2 \\
\\
\underline {Remark}. This unique result does not contradict the results obtained in \\
Example 1, because, there the two asymptotes were those of the relation (1), while the function admits only one.
\\
\\
$y'=\sqrt\frac{x+2}{x-2}+x.\frac{1}{2}\frac{1}{\frac{x+2}{x-2}}\frac{-4}{(x-2)^2}$

\end{exmp}

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